Download Physics 161 Homework 1 - Solutions Friday August 26, 2011

Physics 161
Homework 1 - Solutions
Friday August 26, 2011
Make sure your name is on every page, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them.
The homework is due at the beginning of class on Wednesday, August 31st. Because the
solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. A parsec is defined as the distance such that one second of arc subtends a distance of
one astronomical unit (AU). Find this distance both in meters and in light years.
————————————————————————————————————
Solution
Consider the triangle seen in the figure to the right. The angle is one
π
radisecond of arc, which is 648000
ans, and is tiny such that
1 AU
1 AU ≈ 1 pc × θ.
Now, one astronomical unit is
θ
1 pc
1 AU ≈ 1.496 × 1011 meters.
Thus, one parsec is
1 pc =
1 AU
1.496 × 1011 × 648000
=
= 3.09 × 1016 meters.
θ
π
Now, one light year is the distance light travels in one year, which is 9.46 × 109 meters,
and so one parsec is 3.26 light years.
2. Quantum field theory predicts that forces transmitted by massive particles (as in the
~
, where
weak force) fall off exponentially fast. The effective range of the force is r0 ∼ mc
m is the mass of the messenger particle. The Earth has a static magnetic field that
extends out at least 70,000 km. Using this value, determine the experimental upper
limit for the photon mass, expressed as a fraction of the electron mass.
————————————————————————————————————
Solution
~
~c
The range is r0 = mc
, which we can rewrite as r0 = mc
2 . We did this to use some
handy numbers (which isn’t necessary). It turns out that ~c = 1970 eVÅ, where eV is
an electron volt, and Å = 10−10 meters is an angstrom. Furthermore, mc2 is the rest
mass energy of a mass m. So, we can rewrite
mγ c2 =
~c
.
r0
Now, the range is at least 70,000 km = 7 × 105 meters, or 7 × 1015 Å. Plugging in the
values gives
1970 eVÅ
~c
=
mγ c2 =
= 2.81 × 10−13 eV.
r0
7 × 1015
Now, the rest mass energy of the electron is me c2 = 0.511 MeV = 0.511 × 106 MeV.
−13
Thus, mγ /me = 2.81×10
= 5.5 × 10−19 ∼ 10−18 , where we estimate because of the
0.511×106
approximations used for the range. Thus, the upper bound is
mγ . 10−18 me ,
which is a tiny mass!
3. Confirm the claim that the ratio of gravitational to electrical forces, FG /FE ∼ 10−36
for two protons.
————————————————————————————————————
Solution
The electrical repulsive force between two protons is given by the familiar Coulomb
law,
1 e2
FE =
,
4π0 r2
where r is the distance between the two protons, while the attractive gravitational
force is
m2p
FG = GN 2 ,
r
with mp the mass of the proton. The ratio is, therefore,
m2p
FG
= 4π0 GN 2 .
FE
e
We just need to plug in the numbers,
2
FG
(1.672 × 10−27 )
−37
= 4π × 8.854 × 10−12 × 6.672 × 10−11
∼ 10−36 ,
2 = 8.1 × 10
−19
FE
(1.602 × 10 )
as claimed.
4. In 1899 Max Planck suggested that the constants ~, c, and GN could be combined
in different ways to form natural values of mass, energy, length, and time. Using
dimensional analysis, determine the Planck mass mPl , energy EPl , length `Pl , and time
tPl . Determine the numerical value of each constant.
————————————————————————————————————
Solution
The dimensions of ~ are E · T , or energy times time, while c has dimensions of L/T ,
length per time. Finally, GN has units of E · L/M 2 , where M is mass. Now, to
determine the Planck mass we can write
r
~c
≈ 2.176 × 10−8 kg,
mPl =
GN
which is also 1.22 × 10−19 GeV/c2 . More commonly there is an extra factor of 8π,
r
~c
≈ 2.43 × 1018 GeV/c2 .
mPl ≡
8πGN
Recalling that E = mc2 , we can just multiply the Planck mass by c2 to get the Planck
energy,
s
~c5
EPl = mPl c2 ≡
≈ 2.43 × 1018 GeV.
8πGN
Next, to get the Planck length we again just look at the dimensions of our constants,
writing
r
~GN
≈ 1.616 × 10−35 meters.
`Pl ≡
c3
Finally to get the time we remember that length is velocity times time, so we just
divide the Planck length by the speed of light,
r
`Pl
~GN
tPl =
≡
≈ 5.391 × 10−44 seconds.
5
c
c
5. Explain why Maxwell’s equations explicitly forbid magnetic monopoles.
————————————————————————————————————
Solution
From Maxwell’s equations, the Gauss law for magnetism reads
~ = 0,
∇·B
which may be compared to the equivalent expression for the electric field,
~ =
∇·E
ρ
0
where ρ is the electric charge density. This charge density arises from an excess of
electric charge within some region, including from a single electric charge. Since the
magnetic case is identically zero, that means that there cannot be an excess of magnetic
charge, not even a single magnetic charge. Thus, Maxwell’s equations say that there
are no magnetic monopoles.
6. One of many possible reactions between the electron and positron is e+ + e− → γ + γ,
where the electron and positron annihilate producing two gamma ray photons. Explain
why there must be two photons (hint: suppose that the electron and positron initially
started right next to each other, at rest). What is the energy of each photon, in this
(initially at rest) case?
————————————————————————————————————
Solution
Suppose we had an electron and a positron at rest and just brought them together. In
this case the initial momentum is zero, but the photon carries momentum. So, a single
photon would violate momentum conservation. Two photons, emitted back-to-back,
would conserve momentum. This means we need two photons.
The photons are generated from the annihilation of the two particles, each of mass m.
So, the total initial energy is E = 2mc2 , which goes completely into the two photons.
So, each photon has energy Eγ = me c2 , which is 0.511 MeV.
7. Hubble’s law says that the recessional velocity of distant galaxies is proportional to the
distance away, ~v = H0~r, where H0 is the proportionality constant, called the Hubble
constant. Show that Hubble’s law does not say that we are in the center of the Universe!
(Hint: consider three galaxies, and apply Hubble’s law to each of them, from the point
of view of the others.)
————————————————————————————————————
Solution
The three galaxies, A, B, and C, are seen
in the diagram to the right. According to
Hubble’s law, galaxy A sees galaxy B moving away at speed
vAB
B
~vAB = H(t)~rAB ,
while galaxy A sees galaxy C moving away
at speed
~vAC = H(t)~rAC .
Now, we ask how galaxy C looks from
galaxy B. From the diagram, and since the
velocity is proportional to distance,
rBC
rAB
C
vAC
vAB
vBC
rAC
A
~vBC = ~vAC − ~vAB .
From Hubble’s law we can write
~vBC = H(t)~rAC − H(t)~rAB = H(t) (~rAC − ~rAB ) = H(t)~rBC ,
which tells us that galaxy B sees galaxy C moving away according to Hubble’s law.
Since this is true for any other galaxy, this means that all observers see the same
expansion.
8. The currently accepted value of Hubble’s constant is H0 ≈ 70 km/s/Mpc. Explain how
this value can be used to estimate the age of the Universe, and then do so. Compare
your answer to the current estimate of 13.7 billion years.
————————————————————————————————————
Solution
Hubbles’s law says that v = H0 d, so if we assume that the Universe has been expanding
at a constant speed throughout its lifetime, then the time that the Universe has been
expanding is, roughly,
1
d
.
t∼ =
v
H0
Thus, the Hubble constant has units of inverse time, and gives the overall timescale
associated with the expansion of the Universe. Now, recalling that one parsec is 1.496×
108 kilometers, we can convert
H0 = 70 km/s/Mpc =
70
≈ 2.3 × 10−18 s−1 .
3.09 × 1013 × 106
Thus, we can approximate
T ∼
1
1
≈
≈ 4.41 × 1017 seconds = 13.9 billion years,
H0
2.3 × 10−18
which is very close to the currently accepted value of 13.7 billion years.
9. The net effect of a Cosmological Constant is to provide a constant repulsive force,
becoming important only on megaparsec scales. Explain what effect dark energy would
have on an unstretched spring. What effect would it have on atoms (make a better
estimate than “no effect at all!”)?
————————————————————————————————————
Solution
The Cosmological Constant (CC) acts like a repulsive gravitational force that becomes
big only on large scales. So, on very small scales (say the size of a spring), the net
effect would be to add an additional repulsive force, causing the spring to stretch very
slightly. The electrical forces holding the spring together are vastly larger than the CC
on this scale, and so the only effect would be to shift the equilibrium position by an
infinitesimal amount. The same thing holds true for an atom, where the CC would
cause a completely negligible shift in their equilibrium position, since the electrical
forces completely dominate the CC on these scales.
10. Using the equipartition theorem, show that room temperature corresponds to an average kinetic energy of about 1/40 eV. What temperature (in Kelvins) is high enough
to ionize hydrogen?
————————————————————————————————————
Solution
The equipartition theorem says that every degree of freedom has an average kinetic
energy of E ∼ kB T . So, at room temperature, T ≈ 300 Kelvins, then the energy is
E ∼ kB T = 1.38 × 10−23 (300) = 4.14 × 10−21 Joules,
which, since 1 eV is 1.602 × 10−19 Joules, then the energy is about 0.026 eV, or about
1/40 eV.
The ionization energy of hydrogen is about 10 eV, and so the temperature would need
to be about 400 times bigger than room temperature, or about 120000 K. These are
only estimates, but they are very useful for getting an idea of the numbers involved.